Name the intersection of planes abfe and bfcd. don't mind the selection i made. that is wrong show your work. most answer gets marked brainliest

Two planes intersect in a line that contain all common points of these two planes.

1. Plane named ABFE contains points A, B, F and E.

2. Plane named BFCD contains points B, F, C and D.

As you can see points B and F are common points for these two planes, then the line of intersection is BF and its notation is

Note that notation stays for segments and notation stays for rays.

correct choice is A.

Please see explanation below.

Explanation:

a. Cycle time = Production time available per hour / Units required per hour

= 60 / 5

= 12minutes

Minimum number of workstations = Sum of the task time / Cycle time

Sum of task time

= 8 + 10 + 8 + 10 + 8 + 4

= 48

The theoretical minimum number of work stations is

= 48 / 12

= 4

b. In order to assign the tasks to the work station, events that precede the task must be considered together with the time taken to complete each task.

°Task A This task is assigned to work station 1 and no task would further be assigned to work station 1, otherwise it will exceed the cycle time.

°Task B. This next task will be assigned to work station 2, no additional task will be assigned to station 2.

Task C is assigned to workstation 3, hence can no longer accept any other assigned task.

°Task D is the next task and will be assigned to work station 4, and we cannot assign any more task to work station 4.

°Task E and F will not be assigned as there are no more available stations.

Task Time Workstation

A. 8 1

B. 10 2

C. 8 3

D. 10 4

E. 8 -

F. 4 -

Please note that due to the theoretical minimum number of work station, which is 4, it will not be possible to assign task to all the workstations hence task E and F remains unassigned.

C. Efficiency of the assembly line

Efficiency ;

= Sum of task times / Actual number of work stations × cycle time

Although the actual number of required workstation is 5 but we cannot assign task E and F due to the theoretical minimum number of workstation. Therefore, additional work station will be required and there are 5 work stations in total.

= 48 ÷ (5 × 12) × 100

= 80%